I believe the following ridiculously short function will solve any solvable sextic: Ssexy[q_] := Catch[Block[{rts = #[[1, 2]] & /@ NSolve[q == 0, WorkingPrecision -> 666], foos}, foos = InverseFourier[ rts[[#]]/Sqrt[6]]^6 & /@ (Join[{1}, #] & /@ (1 + Permutations[Range[5]])); Factor[Rationalize[ Expand[Times @@ (# - Rest[foos[[ Ordering[Denominator[Rationalize[Plus @@ #, 9.^-666]] & /@ foos, 1][[1]]]]])]]] /. (a_: 1)*# + b_ :> Throw[Solve[0 == Factor[q, Extension -> Sqrt[-b/a]]]]; Solve[0 == q]]] This is based on the empirical observation that the resolvent quintic, when it exists, will have a linear factor whose root, when square rooted, will split the sextic into cubics. When there is no quintic with rational coeffs, Solve still has a chance of factoring or decomposing. Here's an example of splitting. In[1325]:= MinimalPolynomial[Cos[\[Pi]/5] - Cos[\[Pi]/7], c] Out[1325]= -1 - 8 c + 36 c^2 + 40 c^3 - 128 c^4 - 32 c^5 + 64 c^6 In[1326]:= Ssexy[%] Out[1326]= {{c -> 1/12 (1 - 3 Sqrt[5] + (2 7^(2/3))/(1/2 (1 + 3 I Sqrt[3]))^(1/3) + 2^(2/3) (7 (1 + 3 I Sqrt[3]))^(1/3))}, {c -> 1/12 (1 + 3 Sqrt[5] + (2 7^(2/3))/(1/2 (1 + 3 I Sqrt[3]))^(1/3) + 2^(2/3) (7 (1 + 3 I Sqrt[3]))^(1/3))}, {c -> -((7^(2/3) (1 + I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 - I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) + 1/12 (1 - 3 Sqrt[5])}, {c -> -((7^(2/3) (1 - I Sqrt[3]))/( 6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 + I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) +1/12 (1 - 3 Sqrt[5])}, {c -> -((7^(2/3) (1 + I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 - I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) + 1/12 (1 + 3 Sqrt[5])}, {c -> -((7^(2/3) (1 - I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 + I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) +1/12 (1 + 3 Sqrt[5])}} In[1332]:= FindIntegerNullVector[ Append[#[[1, 2]] & /@ %1326, Cos[\[Pi]/5] - Cos[\[Pi]/7]]] Out[1332]= {0, 0, 0, 0, 1, 0, -1} So Cos[\[Pi]/5] - Cos[\[Pi]/7] is the penultimate solution. --rwg